How to create a filtered DataFrame with minimal code

Question

There are four machines: bmw, geo, vwand porsche:

import pandas as pd
df = pd.DataFrame({
    'car':      ['bmw','geo','vw','porsche'],
    'warranty': ['yes','yes','yes','no'], 
    'dvd':      ['yes','yes','no','yes'], 
    'sunroof':  ['yes','no','no','no']})

I would like to create a filtered DataFrame that lists only those cars that have all three functions: a DVD player, a sunroof and a guarantee (we know that there is a BMW here that has all the functions set to yes).

I can make one column in time with:

cars_with_warranty = df['car'][df['warranty']=='yes']
print(cars_with_warranty)

Then I need to do a similar column calculation for the dvd and sunroof columns:

cars_with_dvd = df['car'][df['dvd']=='yes']
cars_with_sunroof = df['car'][df['sunroof']=='yes']

I wonder if there is a smart way to create a filtered one DataFrame?

EDITED LATER:

. cars_with_all_three - . DataFrame "bmw" , : dvd, sunroof ( , "" ).

cars_with_all_three = []
for ind, car in enumerate(df['car']):
    if df['dvd'][ind] == df['warranty'][ind] == df['sunroof'][ind] == 'yes':
        cars_with_all_three.append(car)

Answer 1

boolean indexing:

print ((df.dvd == 'yes') & (df.sunroof == 'yes') & (df.warranty == 'yes'))
0     True
1    False
2    False
3    False
dtype: bool

print (df[(df.dvd == 'yes') & (df.sunroof == 'yes') & (df.warranty == 'yes')])
   car  dvd sunroof warranty
0  bmw  yes     yes      yes

#if need filter only column 'car' 
print (df.ix[(df.dvd == 'yes')&(df.sunroof == 'yes')&(df.warranty == 'yes'), 'car'])
0    bmw
Name: car, dtype: object

, yes, , True all:

print ((df[[ u'dvd', u'sunroof', u'warranty']] == "yes").all(axis=1))
0     True
1    False
2    False
3    False
dtype: bool

print (df[(df[[ u'dvd', u'sunroof', u'warranty']] == "yes").all(axis=1)])
   car  dvd sunroof warranty
0  bmw  yes     yes      yes

print (df.ix[(df[[ u'dvd', u'sunroof', u'warranty']] == "yes").all(axis=1), 'car'])
0    bmw
Name: car, dtype: object

, DataFrame 4 , sample:

print (df[(df.set_index('car') == 'yes').all(1).values])
   car  dvd sunroof warranty
0  bmw  yes     yes      yes

In [44]: %timeit ([car for ind, car in enumerate(df['car']) if df['dvd'][ind] == df['warranty'][ind] == df['sunroof'][ind] == 'yes'])
10 loops, best of 3: 120 ms per loop

In [45]: %timeit (df[(df.dvd == 'yes')&(df.sunroof == 'yes')&(df.warranty == 'yes')])
The slowest run took 4.39 times longer than the fastest. This could mean that an intermediate result is being cached.
100 loops, best of 3: 2.09 ms per loop

In [46]: %timeit (df[(df[[ u'dvd', u'sunroof', u'warranty']] == "yes").all(axis=1)])
1000 loops, best of 3: 1.53 ms per loop

In [47]: %timeit (df[(df.ix[:, [u'dvd', u'sunroof', u'warranty']] == "yes").all(axis=1)])
The slowest run took 4.46 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 1.51 ms per loop

In [48]: %timeit (df[(df.set_index('car') == 'yes').all(1).values])
1000 loops, best of 3: 1.64 ms per loop

In [49]: %timeit (mer(df))
The slowest run took 4.17 times longer than the fastest. This could mean that an intermediate result is being cached.
100 loops, best of 3: 3.85 ms per loop

:

df = pd.DataFrame({
    'car':      ['bmw','geo','vw','porsche'],
    'warranty': ['yes','yes','yes','no'], 
    'dvd':      ['yes','yes','no','yes'], 
    'sunroof':  ['yes','no','no','no']})

print (df)
df = pd.concat([df]*1000).reset_index(drop=True)

def mer(df):
    df = df.set_index('car')
    return df[df[[ u'dvd', u'sunroof', u'warranty']] == "yes"].dropna().reset_index()

Answer 2

:

df = df.set_index('car')
df[df[[ u'dvd', u'sunroof', u'warranty']] == "yes"].dropna().reset_index()

 df
   car  dvd sunroof warranty
0  bmw  yes     yes      yes


df = df.set_index('car')
df[df[[ u'dvd', u'sunroof', u'warranty']]== "yes"].dropna().index.values

['bmw']   

Answer 3

loop enumerate:

cars_with_all_three = []
for ind, car in enumerate(df['car']):
    if df['dvd'][ind] == df['warranty'][ind] == df['sunroof'][ind] == 'yes':
        cars_with_all_three.append(car)

print(cars_with_all_three), ['bmw'].

Or, if you want to become really smart and use a single liner, you can do this:

[car for ind, car in enumerate(df['car']) if df['dvd'][ind] == df['warranty'][ind] == df['sunroof'][ind] == 'yes']

Hope this helps